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Split List<String> into small concatenated List<String> java 8 parallel execution

So I have 2000 records of company names. I could like to take first 50 names and concatenate it and save as a string and then append it to a new List .
Does anyone have any idea how can we achieve it using java8 ?
Can this be done using parallelstream api?
Currently I’m iteration over 2k records and appending the data to a string builder . Meanwhile after every 50th count I’m creating a new String builder . After every 50 record I add the string builder content to a list. Finally i get list With all the data.
Example: a1# , a2 till a2000
Final output: LiSt of String with
1st entry —> concatenation of a1 to a50,
2nd entry —> concatenation of a51 to a100
Code:
List<String> bulkEmails ; //Fetched from DB
int count = 50;
List<String> splitEmails = new ArrayList<>(); //Final output
StringBuilder builder = new StringBuilder(); // temp builder
for (String mail : bulkEmails) {
builder.append(mail).append(",");
count++;
//append concatenated 50mails, appends to finalOutput and then resets the counter
if (count == 50) {
splitEmails.add(builder.toString());
builder = new StringBuilder();
count = 0;
}
}
Suggestions are appreciated.

As I've said in the comments, 2K isn't really a massive data.
And there's an issue with executing this task in parallel.
Let's imagine this task of splitting the data into groups of 50 elements is running in parallel somehow and the first worker thread is being assigned with a chunk of data containing 523 elements, the second - 518 elements, and so on. And none of these chunks is a product 50. As a consequence of this, each thread would produce a partial result containing a group of size which differs from 50.
Depending on how you want to deal with such cases, there are different approaches on how to implement this functionality:
Join partial results as is. It implies that the final result would contain an arbitrary number of groups having size in range [1,49]. That is the simplest option to implement, and cheapest to execute. But, note that there could even the case when every resulting group is smaller than 50 be since you're not in control of the splitterator implementation (i.e. you can't dictate what would be large would be a chuck of data which particular thread would be assigned to work with). Regardless how strict/lenient your requirements are, that doesn't sound very nice.
The second option requires reordering the elements. While joining partial results produced by each thread in parallel, we can merge the two last groups produced by every thread to ensure that there would be only one group at most that differs in size from 50 in the final result.
If you're not OK with either joining partial results as is, or reordering the elements, that implies that this task isn't suitable for parallel execution because in case when the first thread produces a partial result containing a group of size smaller than 50 all the groups created by the second thread need to be rearranged. Which results in worse performance in parallel because of doing the same job twice.
The second thing we need to consider is that the operation of creating groups requires maintaining a state. Therefore, the right place for this transformation is inside a collector, where the stream is being consumed and mutable container of the collector gets updated.
Let's start with implementing a Collector which ignores the issue described above and joins partial results as is.
For that we can use static method Collector.of().
public static <T> Collector<T, ?, Deque<List<T>>> getGroupCollector(int groupSize) {
return Collector.of(
ArrayDeque::new,
(Deque<List<T>> deque, T next) -> {
if (deque.isEmpty() || deque.getLast().size() == groupSize) deque.add(new ArrayList<>());
deque.getLast().add(next);
},
(left, right) -> {
left.addAll(right);
return left;
}
);
}
Now let's implement a Collector which merges the two last groups produced in different threads (option 2 in the list above):
public static <T> Collector<T, ?, Deque<List<T>>> getGroupCollector(int groupSize) {
return Collector.of(
ArrayDeque::new,
(Deque<List<T>> deque, T next) -> {
if (deque.isEmpty() || deque.getLast().size() == groupSize) deque.add(new ArrayList<>());
deque.getLast().add(next);
},
(left, right) -> {
if (left.peekLast().size() < groupSize) {
List<T> leftLast = left.pollLast();
List<T> rightLast = right.peekLast();
int llSize = leftLast.size();
int rlSize = rightLast.size();
if (rlSize + llSize <= groupSize) {
rightLast.addAll(leftLast);
} else {
rightLast.addAll(leftLast.subList(0, groupSize - rlSize));
right.add(new ArrayList<>(leftLast.subList(groupSize - rlSize, llSize)));
}
}
left.addAll(right);
return left;
}
);
}
If you wish to implement a Collector which combiner function rearranges the partial results if needed (the very last option in the list), I'm leaving it to the OP/reader as an exercise.
Now let's use the collector defined above (the latest). Let's consider a stream of single-letter strings representing the characters of the English alphabet, and let's group size would be 5.
public static void main(String[] args) {
List<String> groups = IntStream.rangeClosed('A', 'Z')
.mapToObj(ch -> String.valueOf((char) ch))
.collect(getGroupCollector(5))
.stream()
.map(group -> String.join(",", group))
.collect(Collectors.toList());
groups.forEach(System.out::println);
}
Output:
A,B,C,D,E
F,G,H,I,J
K,L,M,N,O
P,Q,R,S,T
U,V,W,X,Y
Z
That's the sequential result. If you switch, the stream from sequential to parallel the contents of the groups would probably change, but the sizes would not be affected.
Also note that since there are effectively two independent streams chained together, you need to apply parrallel() twice to make the whole thing working in parallel.

given an infinite sequence break it into intervals, and return a new infinite sequence with the average of each interval

i have to calculate the average of a Infinite Sequence using Stream API
Input:
Stream<Double> s = a,b,c,d ...
int interval = 3
Expected Result:
Stream<Double> result = avg(a,b,c), avg(d,e,f), ....
the result can be also an Iterator, or any other type
as long as it mantains the structure of an infinite list
of course what i written is pseudo code and doesnt run

There is a #Beta API termed mapWithIndex within Guava that could help here with certain assumption:
static Stream<Double> stepAverage(Stream<Double> stream, int step) {
return Streams.mapWithIndex(stream, (from, index) -> Map.entry(index, from))
.collect(Collectors.groupingBy(e -> (e.getKey() / step), TreeMap::new,
Collectors.averagingDouble(Map.Entry::getValue)))
.values().stream();
}
The assumption that it brings in is detailed in the documentation clearly(emphasized by me):
The resulting stream is efficiently splittable if and only if stream
was efficiently splittable and its underlying spliterator reported
Spliterator.SUBSIZED. This is generally the case if the underlying
stream comes from a data structure supporting efficient indexed random
access, typically an array or list.

This should work fine using vanilla Java
I'm using Stream#mapMulti and a Set external to the Stream to aggregate the doubles
As you see, I also used DoubleSummaryStatistics to count the average.
I could have use the traditional looping and summing then dividing but I found this way more explicit
Update:
I changed the Collection used from Set to List as a Set could cause unexpected behaviour
int step = 3;
List<Double> list = new ArrayList<>();
Stream<Double> averagesStream =
infiniteStream.mapMulti(((Double aDouble, Consumer<Double> doubleConsumer) -> {
list.add(aDouble);
if (list.size() == step) {
DoubleSummaryStatistics doubleSummaryStatistics = new DoubleSummaryStatistics();
list.forEach(doubleSummaryStatistics::accept);
list.clear();
doubleConsumer.accept(doubleSummaryStatistics.getAverage());
}
}));

Accumulating value of objects when carrying the same timestamp

I am currently stuck on this:
I have datapoints that carry a value and a timestamp as a Long (epoch seconds):
public class MyDataPoint(){
private Float value;
private Long timestamp;
//constructor, getters and setters here
}
I have lists that are bound to different sources where these datapoints are coming from.
public class MySource(){
private Interger sourceId;
private List<MyDataPoint> dataPointList;
//constructor, getters and setters here
}
Now I want to accumulate these datapoints in a new list:
each datapoint with the same timestamp should be accumulated in a new datapoint with the sum of the value of each datapoint that carries the same timestamp.
So for instance I have 3 datapoints with the same timestamp, I want to create one datapoint with the timestamp, and the sum of the three values.
However, these datapoints have not started or ended recording at the same time. And for one timestamp maybe only one datapoint exists.
For now I have stuffed all of the datapoints into one list, thinking I could use streams to achieve my goal, but I can't figure it out. Maybe this is the wrong way anyway because I can't see how to use filters or maps to do this.
I have thought about using Optionals since for one timestamp maybe only one exists, but there is no obvious answer for me.
Anyone able to help me out?

I am guessing that you are trying to grouping the value you in the list, then convert it to new list using stream. What i suggest is using Collectors.groupingBy and Collectors.summingInt to convert your List to a Map<Long,Double> first - which holding your timestamp as key and Double as sum of all value that has same timestamp. After this you can convert this map back to the new list.
Not tested yet but to convert your List to Map<Long, Double> should be something like:
dataPointList.stream().collect(Collectors.groupingBy(d -> d.timestamp, Collectors.summingDouble(d -> d.value))); //you can using method reference for better readability

Following assumes your DataPoint is immutable (you cannot use the same instance to accumulate into) so uses an intermediate Map.
Collection<DataPoint> summary = sources.stream()
.flatMap(source -> source.dataPointList.stream()) // smush sources into a single stream of points
.collect(groupingBy(p -> p.timestamp, summingDouble(p -> (double)p.value))) // Collect points into Map<Long, Double>
.entrySet().stream() // New stream, the entries of the Map
.map(e -> new MyDataPoint(e.getKey(), e.getValue()))
.collect(toList());
Another solution avoids the potentially large intermediate Map by collecting directly into a DataPoint.
public static DataPoint combine(DataPoint left, DataPoint right) {
return new DataPoint(left.timestamp, left.value + right.value); // return new if immutable or increase left if not
}
Collection<DataPoint> summary = sources.stream()
.flatMap(source -> source.dataPointList.stream()) // smush into a single stream of points
.collect(groupingBy(p -> p.timestamp, reducing(DataPoint.ZERO, DataPoint::combine))) // Collect all values into Map<Long, DataPoint>
.values();
This can be upgraded to parallelStream() if DataPoint is threadsafe etc

I think the "big picture" solution it's quite easy even if I can predict some multithread issues to complicate all.
In pure Java, you need simply a Map:
Map<Long,List<MyDataPoint>> dataPoints = new HashMap<>();
just use Timestamp as KEY.
For the sake of OOP, Let's create a class like DataPointCollector
public class DataPointCollector {
private Map<Long,List<MyDataPoint>> dataPoints = new HashMap<>();
}
To add element, create a method in DataPointCollector like:
public void addDataPoint(MyDataPoint dp){
if (dataPoints.get(dp.getTimestamp()) == null){
dataPoints.put(dp.getTimestamp(), new ArrayList<MyDataPoint>());
}
dataPoints.get(dp.getTimestamp()).add(dp);
}
This solve most of your theorical problems.
To get the sum, just iterate over the List and sum the values.
If you need a realtime sum, just wrap the List in another object that has totalValue and List<MyDataPoint> as fields and update totalValue on each invokation of addDataPoint(...).
About streams: streams depends by use cases, if in a certain time you have all the DataPoints you need, of course you can use Streams to do things... however streams are often expensive for common cases and I think it's better to focus on an easy solution and then make it cool with streams only if needed

what is the difference between a stateful and a stateless lambda expression?

According to the OCP book one must avoid stateful operations otherwise known as stateful lambda expression. The definition provided in the book is 'a stateful lambda expression is one whose result depends on any state that might change during the execution of a pipeline.'
They provide an example where a parallel stream is used to add a fixed collection of numbers to a synchronized ArrayList using the .map() function.
The order in the arraylist is completely random and this should make one see that a stateful lambda expression produces unpredictable results in runtime. That's why its strongly recommended to avoid stateful operations when using parallel streams so as to remove any potential data side effects.
They don't show a stateless lambda expression that provides a solution to the same problem (adding numbers to a synchronized arraylist) and I still don't get what the problem is with using a map function to populate an empty synchronized arraylist with data... What is exactly the state that might change during the execution of a pipeline? Are they referring to the Arraylist itself? Like when another thread decides to add other data to the ArrayList when the parallelstream is still in the process adding the numbers and thus altering the eventual result?
Maybe someone can provide me with a better example that shows what a stateful lambda expression is and why it should be avoided. That would be very much appreciated.
Thank you

The first problem is this:
List<Integer> list = new ArrayList<>();
List<Integer> result = Stream.of(1, 2, 3, 4, 5, 6)
.parallel()
.map(x -> {
list.add(x);
return x;
})
.collect(Collectors.toList());
System.out.println(list);
You have no idea what the result will be here, since you are adding elements to a non-thread-safe collection ArrayList.
But even if you do:
List<Integer> list = Collections.synchronizedList(new ArrayList<>());
And perform the same operation the list has no predictable order. Multiple Threads add to this synchronized collection. By adding the synchronized collection you guarantee that all elements are added (as opposed to the plain ArrayList), but in which order they will be present in unknown.
Notice that list has no order guarantees what-so-ever, this is called processing order. While result is guaranteed to be: [1, 2, 3, 4, 5, 6] for this particular example.
Depending on the problem, you usually can get rid of the stateful operations; for your example returning the synchronized List would be:
Stream.of(1, 2, 3, 4, 5, 6)
.filter(x -> x > 2) // for example a filter is present
.collect(Collectors.collectingAndThen(Collectors.toList(),
Collections::synchronizedList));

To try to give an example, let's consider the following Consumer (note : the usefulness of such a function is not of the matter here) :
public static class StatefulConsumer implements IntConsumer {
private static final Integer ARBITRARY_THRESHOLD = 10;
private boolean flag = false;
private final List<Integer> list = new ArrayList<>();
#Override
public void accept(int value) {
if(flag){ // exit condition
return;
}
if(value >= ARBITRARY_THRESHOLD){
flag = true;
}
list.add(value);
}
}
It's a consumer that will add items to a List (let's not consider how to get back the list nor the thread safety) and has a flag (to represent the statefulness).
The logic behind this would be that once the threshold has been reached, the consumer should stop adding items.
What your book was trying to say was that because there is no guaranteed order in which the function will have to consume the elements of the Stream, the output is non-deterministic.
Thus, they advise you to only use stateless functions, meaning they will always produce the same result with the same input.

Here is an example where a stateful operation returns a different result each time:
public static void main(String[] args) {
Set<Integer> seen = new HashSet<>();
IntStream stream = IntStream.of(1, 2, 3, 1, 2, 3);
// Stateful lambda expression
IntUnaryOperator mapUniqueLambda = (int i) -> {
if (!seen.contains(i)) {
seen.add(i);
return i;
}
else {
return 0;
}
};
int sum = stream.parallel().map(mapUniqueLambda).peek(i -> System.out.println("Stream member: " + i)).sum();
System.out.println("Sum: " + sum);
}
In my case when I ran the code I got the following output:
Stream member: 1
Stream member: 0
Stream member: 2
Stream member: 3
Stream member: 1
Stream member: 2
Sum: 9
Why did I get 9 as the sum if I'm inserting into a hashset?
The answer: Different threads took different parts of the IntStream
For example values 1 & 2 managed to end up on different threads.

A stateful lambda expression is one whose result depends on any state that might change during the execution of a pipeline. On the
other hand, a stateless lambda expression is one whose result does
not depend on any state that might change during the execution of a
pipeline.
Source: OCP: Oracle Certified Professional Java SE 8 Programmer II Study Guide: Exam 1Z0-809by Jeanne Boyarsky,‎ Scott Selikoff
List < Integer > data = Collections.synchronizedList(new ArrayList < > ());
Arrays.asList(1, 2, 3, 4, 5, 6, 7).parallelStream()
.map(i -> {
data.add(i);
return i;
}) // AVOID STATEFUL LAMBDA EXPRESSIONS!
.forEachOrdered(i -> System.out.print(i+" "));
System.out.println();
for (int e: data) {
System.out.print(e + " ");
Possible Output:
1 2 3 4 5 6 7
1 7 5 2 3 4 6
It strongly recommended that you avoid stateful operations when using
parallel streams, so as to remove any potential data side effects. In
fact, they should generally be avoided in serial streams wherever
possible, since they prevent your streams from taking advantage of
parallelization.

A stateful lambda expression is one whose result depends on any state that might change during the execution of a stream pipeline.
Let's understand this with an example here:
List<Integer> list = Arrays.asList(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15);
List<Integer> result = new ArrayList<Integer>();
list.parallelStream().map(s -> {
synchronized (result) {
if (result.size() < 10) {
result.add(s);
}
}
return s;
}).forEach( e -> {});
System.out.println(result);
When you run this code 5 times, the output would/could be different all the time. Reason behind is here processing of Lambda expression inside map updates result array. Since here the result array depend on the size of that array for a particular sub stream, which would change every time this parallel stream would be called.
For better understanding of parallel stream:
Parallel computing involves dividing a problem into subproblems, solving those problems simultaneously (in parallel, with each subproblem running in a separate thread), and then combining the results of the solutions to the subproblems. When a stream executes in parallel, the Java runtime partitions the streams into multiple substreams. Aggregate operations iterate over and process these substreams in parallel and then combine the results.
Hope this helps!!!

Passing a collection using a reduce (3 parameters) function - streams java 8

I am trying to calculate the multiplication of a value using the previous two values using java 8's stream. I want to call a function that will return an array/list/collection. I am creating a List and adding 1,2 to it.
Let's say the list name is result.
public static void main (String[] args) {
List<Integer> result = new ArrayList<Integer>();
result.add(1);
result.add(2);
int n = 5; //n can be anything, choosing 5 for this example
res(n, result);
//print result which should be [1, 2, 2, 4, 8]
}
public static List<Integer> res(int n, List<Integer> result ) {
result.stream()
.limit(n)
.reduce(identity, (base,index) -> base);
//return result;
}
Now the issue is trying to try to pass result into the stream to keep updating the list with the new values using the stream. According to the java tutorials, it is possible, albeit inefficient.
"If your reduce operation involves adding elements to a collection, then every time your accumulator function processes an element, it creates a new collection that includes the element, which is inefficient."
Do I need to use the optional third parameter, BinaryOperator combiner, to combine the list + result??
<U> U reduce(U identity,
BiFunction<U,? super T,U> accumulator,
BinaryOperator<U> combiner)
In short; I want to pass a list with two values and have the function find the multiplication of the first two values (1,2), add it to the list, and find the multiplication of the last two values (2,2), and add it to the list, and until the stream hits the limit.

It looks like you're trying to implement a recurrence relation. The reduce method applies some function to a bunch of pre-existing values in the stream. You can't use reduce and take an intermediate result from the reducer function and "feed it back" into the stream, which is what you need to do in order to implement a recurrence relation.
The way to implement a recurrence relation using streams is to use one of the streams factory methods Stream.generate or Stream.iterate. The iterate factory seems to suggest the most obvious approach. The state that needs to be kept for each application of the recurrence function requires two ints in your example, so unfortunately we have to create an object to hold these for us:
static class IntPair {
final int a, b;
IntPair(int a_, int b_) {
a = a_; b = b_;
}
}
Using this state object you can create a stream that implements the recurrence that you want:
Stream.iterate(new IntPair(1, 2), p -> new IntPair(p.b, p.a * p.b))
Once you have such a stream, it's a simple matter to collect the values into a list:
List<Integer> output =
Stream.iterate(new IntPair(1, 2), p -> new IntPair(p.b, p.a * p.b))
.limit(5)
.map(pair -> pair.a)
.collect(Collectors.toList());
System.out.println(output);
[1, 2, 2, 4, 8]
As an aside, you can use the same technique to generate the Fibonacci sequence. All you do is provide a different starting value and iteration function:
Stream.iterate(new IntPair(0, 1), p -> new IntPair(p.b, p.a + p.b))
You could also implement a similar recurrence relation using Stream.generate. This will also require a helper class. The helper class implements Supplier of the result value but it also needs to maintain state. It thus needs to be mutable, which is kind of gross in my book. The iteration function also needs to be baked into the generator object. This makes it less flexible than the IntPair object, which can be used for creating arbitrary recurrences.

Just for completeness, here is a solution which does not need an additional class.
List<Integer> output = Stream.iterate(
(ToIntFunction<IntBinaryOperator>)f -> f.applyAsInt(1, 2),
prev -> f -> prev.applyAsInt((a, b) -> f.applyAsInt(b, a*b) )
)
.limit(9).map(pair -> pair.applyAsInt((a, b)->a))
.collect(Collectors.toList());
This is a functional approach which doesn’t need an intermediate value storage. However, since Java is not a functional programming language and doesn’t have optimizations for such a recursive function definition, this is not recommended for larger streams.
Since for this example a larger stream would overflow numerically anyway and the calculation is cheap, this approach works. But for other use cases you will surely prefer a storage object when solving such a problem with plain Java (as in Stuart Marks’ answer)